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Full text of "Scientific Papers - Vi"

^5]      HYDKODYNAMICAL  PROBLEMS  SUGGESTED  BY PITOT'S TUBES         333
,nd w (= <jb + i/v|r) is expressed with the aid of an auxiliary variable Q.
us
z = tan 6— 6 — ;^'tan2 0 - i log cos 6,   .......'...........(6)
1           0  /]                                                                         / ^7 \
ni\ —r   _L   Qpp* ri                                                                              I   /   1
tJU ~u>   7T   *-3C\_;    (/•      .*.**••••*<•>. !•••. <i*t«*. .»>.»•.Ill
If we put tan 8 = f -I-177, we get
that                        <H
We find further (Love),
(8)
= | + Mj
that
a; = £ + i/r + | tan"1 —
+
|2 _
.........(9)
,   ......(10)
The stream -lines, corresponding to a constant -fy, may be plotted from )), (11), if we substitute 2^/£ for T? and regard £ as the variable parameter. ice by (8)
0 = i (1 + f9) - tY£2>      <ty/# = if + 2^Vf,
>re is no occasion to consider negative values of f, and 0 arid f vary always the same direction.
As regards the fractions under the sign of tan~a, we see that both vanish en £ = 0, and also when £=oo. The former, viz., 2£-5-(4nJra/ffa + f-1), first + when £ is very small, rises to oc when £2 = £{1 ± \/(l -16^2)}y ich happens when ^ < \, but not otherwise. In the latter case the fraction always positive. When -^ < | , the fraction passes through oo , there inging sign. The numerically least, negative value is reached when = £|Y(1 + 48^2) — 1}. The fraction then retraces its entire course, until becomes zero again when £ = oo . On the other hand the second fraction,. first positive, rises to infinity in all cases when |2= i |\/ (1 + 16-^2) — 1}, er which it becomes negative and decreases numerically to zero, no part of course being retraced. As regards the ambiguities in the resulting angles,. will suffice to suppose both angles to start from zero with £. This choice Lounts to taking the origin of oc at 0, instead of 0'.
When A/T is very small the march of the functions is peculiar. The first. ction becomes infinite when £3 = 4Ajr2, that is when £ is still small. The rn occurs when £2 = 12ip, and the corresponding least negative value is o small. The first tan"1 thus passes from 0 to TT while £ is still small. ie second fraction also becomes infinite when £2 = 4i/r2, there changing ;n, and again approaches zero while £ is of the same order of magnitude.ened.                                                                                                                            *